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杭电 1003

2024-07-14 13:37:47   226人阅读

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 142647    Accepted Submission(s): 33192


Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
 


 

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
 


 

Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
 


 

Sample Input
25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5
 


 

Sample Output
Case 1:14 1 4Case 2:7 1 6
 


 

Author
Ignatius.L

 

这道题的题目虽然是大数 但是不用大数也是能够做得出来的

题目的意思就是给你一段数字，开头还有结尾都是可以变化的，

在两者都不确定的情况下，让你求最长和的子列，并且将最

大的值，子列的起始位置，终止位置给输出来；

这是简单的dp问题

 

 

代码如下：

#include<stdio.h>
int main()
{
 int n;
 int max,i,j,m,start,end,k,temp,a;
 scanf("%d",&n);
 for(i=0;i<n;i++)//此处写成while(n--)也行 但是后面对case值的变化 要多一个变量进行描述
 {
  max=-1001;
  end=start=k=temp=0;
  scanf("%d",&m);
  for(j=0;j<m;j++)
  {
   scanf("%d",&a);
   temp+=a;//temp起到了缓存的效果，将总和的值加起来之后再与最大值比较
   if(temp>max)
   {
    max=temp;
    start=k;
    end=j;
   }
   if(temp<0)//此处要是前面的和小于零 说明 不可能从其那面开始 所以其实位置也要发生变化
   {
    temp=0;
    k=j+1;
   }
  }
  printf("Case %d:\n",i+1);
  printf("%d %d %d\n",max,start+1,end+1);
  if(i!=n-1)
  printf("\n");
 }
 return 0;
}


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